A STUDY OF BANK CUSTOMER CHURN

Description of each variable:

• Credit_score — credit rating
• geography — client country
• gender — client gender
• age — client age
• tenure — number of years spent by the client with the bank
• balance — client account balance
• num_of_product — number of products that the client purchased from the bank
• has_cr_card — the client has a credit card
• is-_active_member — active client
• estimated_salary — clien salary
• exited — client left the bank

# load libraries
import pandas as pd
import numpy as np
import seaborn as sns
from matplotlib import pyplot as plt   #plt.xkcd()
from IPython.display import set_matplotlib_formats
import matplotlib.ticker as mtick
from sklearn.preprocessing import LabelEncoder
from utilities import *
import scipy.stats as ss
from scipy.stats import ttest_ind
from sklearn.metrics import confusion_matrix
from sklearn.preprocessing import StandardScaler
from sklearn.model_selection import train_test_split
from itertools import combinations
import warnings
warnings.filterwarnings("ignore")

# Improve quality of the plots
set_matplotlib_formats('svg')
# Set the style of the plot
plt.style.use('seaborn-v0_8-white')
#plt.xkcd()

1-Load and clean the data

# Load the data
df = pd.read_csv("churn.csv", index_col=False)
df.head(5)

	RowNumber	CustomerId	Surname	CreditScore	Geography	Gender	Age	Tenure	Balance	NumOfProducts	HasCrCard	IsActiveMember	EstimatedSalary	Exited
0	1	15634602	Hargrave	619	France	Female	42	2	0.00	1	1	1	101348.88	1
1	2	15647311	Hill	608	Spain	Female	41	1	83807.86	1	0	1	112542.58	0
2	3	15619304	Onio	502	France	Female	42	8	159660.80	3	1	0	113931.57	1
3	4	15701354	Boni	699	France	Female	39	1	0.00	2	0	0	93826.63	0
4	5	15737888	Mitchell	850	Spain	Female	43	2	125510.82	1	1	1	79084.10	0

• At least 3 columns do not provide training value : RowNumber, CustomerId and Surname

col_name = ['credit_score', 'geography', 'gender', 'age', 'tenure', 'balance', 
            'num_of_products', 'has_cr_card', 'is_active_member', 'estimated_salary', 'exited']

# Remove unnecessary variables
df = df.drop(["RowNumber", "CustomerId", "Surname"], axis=1)
# Change the columns name
df.columns = col_name
# Show the data
df.head(5)

	credit_score	geography	gender	age	tenure	balance	num_of_products	has_cr_card	is_active_member	estimated_salary	exited
0	619	France	Female	42	2	0.00	1	1	1	101348.88	1
1	608	Spain	Female	41	1	83807.86	1	0	1	112542.58	0
2	502	France	Female	42	8	159660.80	3	1	0	113931.57	1
3	699	France	Female	39	1	0.00	2	0	0	93826.63	0
4	850	Spain	Female	43	2	125510.82	1	1	1	79084.10	0

df.info()

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 10000 entries, 0 to 9999
Data columns (total 11 columns):
 #   Column            Non-Null Count  Dtype  
---  ------            --------------  -----  
 0   credit_score      10000 non-null  int64  
 1   geography         10000 non-null  object 
 2   gender            10000 non-null  object 
 3   age               10000 non-null  int64  
 4   tenure            10000 non-null  int64  
 5   balance           10000 non-null  float64
 6   num_of_products   10000 non-null  int64  
 7   has_cr_card       10000 non-null  int64  
 8   is_active_member  10000 non-null  int64  
 9   estimated_salary  10000 non-null  float64
 10  exited            10000 non-null  int64  
dtypes: float64(2), int64(7), object(2)
memory usage: 859.5+ KB

• The set contains information about 10,000 bank users

• No empty values found

• There are both nominative and quantitative characteristics

# Number of unique value for each variable
df.nunique()

credit_score         460
geography              3
gender                 2
age                   70
tenure                11
balance             6382
num_of_products        4
has_cr_card            2
is_active_member       2
estimated_salary    9999
exited                 2
dtype: int64

2-Statistics and data visualisation

Bar plot of binary variables

plt.figure(figsize=(8, 8))
bar_columns = ["exited", "gender", "has_cr_card", "is_active_member"]
counter = 1
for key in bar_columns:
    plt.subplot(2, 2, counter)
    bar_plot(df[key], x_label=key, y_label="%", color=['indigo', 'gold'])
    counter += 1

• Gender and activities among customers are evenly distributed.
• The proportion of customers holding a credit card is lower.
• The proportion of exiting customers is low.

Box plot of continuous variables

numeric_col = ['age', 'estimated_salary', 'balance', 'credit_score']

plt.figure(figsize=(10, 10))
counter = 1
for key in numeric_col:
    plt.subplot(2, 2, counter)
    sns.boxplot(x = df['exited'], y = df[key], palette=['indigo', 'gold'])\
                                                    .set(title=f'{key} by exited')
    counter += 1

Correlation between numerical variables

# Correlation between numrical variable
plt.figure(figsize=(9, 6))
sns.heatmap(df.drop(['geography', 'gender'], axis=1).corr(), annot=True, cmap='viridis', linewidths=.5)
plt.title('Correlation Plot')

Text(0.5, 1.0, 'Correlation Plot')

• Based on the chart provided, there is no strong linear relationship observed between the variables.

Association between categorical variables

For this purpose, we will use Cramer's V. It measures how strongly two categorical fields are associated.

$$ \mathcal{V} = \sqrt{\frac{\chi^2/n}{min(r-1, c-1)}} $$

categorical_col = ["num_of_products", "gender", "has_cr_card", "is_active_member", "geography"]
score = {}
exited = df["exited"]
for col in categorical_col:
    temp = df[col]
    the_confusion_matrix = pd.crosstab(exited, temp).to_numpy()
    score[col] = CramerV(the_confusion_matrix, True)

score = pd.Series(score, index=categorical_col).sort_values(ascending=False)

score.plot(kind="bar", grid=True, color=['#4B0082', '#4B008266', '#4B008299', '#4B0082CC'])
plt.xticks(rotation=50, ha='right')
plt.title('Exited and Categorical')

Text(0.5, 1.0, 'Exited and Categorical')

• A Cramer's V value of 0.4 indicates a moderate level of association between exited and num_of_products. The association is more substantial compared to the previous value, but it's still not very strong.

• The remaine variables suggests a relatively weak association with the exited variable.

Conditional probabilities

#--------------------------------------------
# Calculating the continitional probabilities 
#--------------------------------------------
probs = {}
for col in categorical_col:
    uniques = df[col].unique()
    for unique in uniques:
        condition = df[col] == unique
        probs[f'P(Exited|{col}={unique})'] = round(df.loc[condition, 'exited'].mean(), 2)


probs = pd.Series(probs).sort_values(ascending=False)

#--------------------------------------------
# plot the  probabilities 
#--------------------------------------------
colors = ["#FF0000", "#FF4500", "#FFA500", "#FFD700", "#FFFF00","#ADFF2F", "#7FFF00", "#00FF00", 
          "#00FA9A", "#00CED1", "#4682B4", "#0000FF", "#4B0082"]
probs.plot(kind="bar", grid = True, color = colors)
plt.xticks(rotation=50, ha='right')
plt.title('Probability of Exited given Categorical')

Text(0.5, 1.0, 'Probability of Exited given Categorical')

• Customers with 4 or 3 products face a higher risk of exiting the bank compared to those with 1 or 2 products.

• For the other variables, the risk of exiting is significantly low.

Distribution of continuous variables

fig, axes = plt.subplots(2,2, figsize = (9,9))
plt.subplots_adjust(hspace=0.5)
columns = df[numeric_col]
for i, column in enumerate(columns):
    ax = axes[i // 2, i % 2]
    sns.kdeplot(data = df,
                x = column,
                fill = True,
                alpha = 0.5,
                hue = 'exited',
                palette = ['#6A5ACD', '#4B0082'],
                ax = ax)
    ax.set_xlabel(column, fontsize = 14)
plt.show()

• Given the distributions above, we observe that the distribution changes when clients exit or stay.
• We observe a heavy tail in the distributions, which implies higher probabilities for the extreme values.
• Clients over 70 and below 20 have a higher chance of not exiting..

plt.figure(figsize=(10, 9))
paired_list_numerical = list(combinations(numeric_col, 2))
counter = 1
for key in paired_list_numerical:
    plt.subplot(3, 2, counter)
    sns.scatterplot(x=key[0], y=key[1], hue="exited", palette=['#6A5ACD', "gold"], data=df) 
    counter += 1

• According to the depicted plot, as customer age increases, there appears to be a tendency for the balance to converge towards the median value. Otherwie, there is no significant relationship discernible among the other numerical variables.

fig, axes = plt.subplots(2,2, figsize = (9,9))
plt.subplots_adjust(hspace=0.5)
columns = df[numeric_col]
for i, column in enumerate(columns):
    ax = axes[i // 2, i % 2]
    sns.kdeplot(data = df,
                x = column,
                fill = True,
                alpha = 0.5,
                hue = 'gender',
                palette = ['#4B0082', '#6A5ACD'],
                ax = ax)
    ax.set_xlabel(column, fontsize = 14)
plt.show()

We observe slightly higher values in credit score, estimated salary, age, and account balance for males compared to females. This difference is linked to the greater number of male customers as opposed to female customers.

fig, axes = plt.subplots(2,2, figsize = (10,10))
plt.subplots_adjust(hspace=0.5)
columns = df[numeric_col]
for i, column in enumerate(columns):
    ax = axes[i // 2, i % 2]
    sns.kdeplot(data = df,
                x = column,
                fill = True,
                alpha = 0.5,
                hue = 'geography',
                palette = ['#17b3a8', '#ff0a89', '#4682B4'],
                ax = ax)
    ax.set_xlabel(column, fontsize = 14)
plt.show()

• For age, credit score, balance, and estimated salary, France exhibits higher values in the tail, indicating a higher probability of extreme values.

Summary of continuous variables

# Summary of numerical variables
df[numeric_col].describe()

	age	estimated_salary	balance	credit_score
count	10000.000000	10000.000000	10000.000000	10000.000000
mean	38.921800	100090.239881	76485.889288	650.528800
std	10.487806	57510.492818	62397.405202	96.653299
min	18.000000	11.580000	0.000000	350.000000
25%	32.000000	51002.110000	0.000000	584.000000
50%	37.000000	100193.915000	97198.540000	652.000000
75%	44.000000	149388.247500	127644.240000	718.000000
max	92.000000	199992.480000	250898.090000	850.000000

Mean of continuous variables given the exited variable

ex_cr = df.groupby('exited')['credit_score'].mean().reset_index()
ex_age = df.groupby('exited')['age'].mean().reset_index()
merge_1 = pd.merge(ex_cr, ex_age, on='exited')
ex_salary = df.groupby('exited')['estimated_salary'].mean().reset_index()
merge_2 = pd.merge(merge_1, ex_salary, on='exited')
ex_balance = df.groupby('exited')['estimated_salary'].mean().reset_index()
pd.merge(merge_2, ex_balance, on='exited')

	exited	credit_score	age	estimated_salary_x	estimated_salary_y
0	0	651.853196	37.408389	99738.391772	99738.391772
1	1	645.351497	44.837997	101465.677531	101465.677531

Visually, the conditional mean of continuous variables shows minimal distinctions.

To validate this observation, we will conduct statistical tests.

We will test if

$$ H_0 : \mathbb{E}[X_i|exited = 1] - \mathbb{E}[X_i|exited = 0] = 0 $$

$$ H_1 : \mathbb{E}[X_i|exited = 1] - \mathbb{E}[X_i|exited = 0] \neq 0 $$

$H_0$ : No significant difference between the means

$H_1$ : There is a significant difference between the means

The sample mean : $\bar{Y}_{1} = \frac{1}{n_1}\sum_{i=1}^{n_1}Y_{1, i}$ and $\bar{Y}_{2} = \frac{1}{n_2}\sum_{i=1}^{n_2}$

The sample variance : $S^2_1 = \frac{1}{n_1-1}\sum_{i=1}^{n_1}(Y_{1,i} - \bar{Y}_{1})^2$ and $S^2_2 = \frac{1}{n_2-1}\sum_{i=1}^{n_2}(Y_{2,i} - \bar{Y}_{2})^2$

$S^2_p = \frac{(n_1 - 1)S^2_1 + (n_2 - 1)S^2_2}{n_1 + n_2 - 2}$

$S^2_p$ : is the common variance estimation

We calculate the t value :

$$ t = \frac{\bar{Y}_{1} - \bar{Y}_{2}}{S_p\sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} $$

Finaly, after setting a level $\alpha$, we can calcula the p_value. If the $p-value > \alpha$ we don't reject $H_0$, otherwise, we rejected.

$$ p-value = \mathbb{P}(T > |t|) $$

where

$$ T \sim \mathcal{Student}(n_1 + n_2 -2) $$

student_test(df, numeric_col, group_by='exited', alpha_level=0.05)

[age, exited] : Reject the null hypothesis with level 0.05; there is a significant difference between groups.
===
[estimated_salary, exited] : Fail to reject the null hypothesis with level 0.05; there is no significant difference between groups.
===
[balance, exited] : Reject the null hypothesis with level 0.05; there is a significant difference between groups.
===
[credit_score, exited] : Reject the null hypothesis with level 0.05; there is a significant difference between groups.
===

• There is no significant difference in the mean of estimated_salary between individuals who stay and those who exit.

• For the other variables, according to the t-test results, there is a significant difference.

Mean of continuous variables given the gender

student_test(df, numeric_col, group_by='gender', alpha_level=0.05)

[age, gender] : Reject the null hypothesis with level 0.05; there is a significant difference between groups.
===
[estimated_salary, gender] : Fail to reject the null hypothesis with level 0.05; there is no significant difference between groups.
===
[balance, gender] : Fail to reject the null hypothesis with level 0.05; there is no significant difference between groups.
===
[credit_score, gender] : Fail to reject the null hypothesis with level 0.05; there is no significant difference between groups.
===

• There is significant difference in the mean of age between male and female.

• For the other variables, according to the t-test results, there is no significant difference.

Mean of continuous variables given the is active member

student_test(df, numeric_col, group_by='is_active_member', alpha_level=0.05)

[age, is_active_member] : Reject the null hypothesis with level 0.05; there is a significant difference between groups.
===
[estimated_salary, is_active_member] : Fail to reject the null hypothesis with level 0.05; there is no significant difference between groups.
===
[balance, is_active_member] : Fail to reject the null hypothesis with level 0.05; there is no significant difference between groups.
===
[credit_score, is_active_member] : Reject the null hypothesis with level 0.05; there is a significant difference between groups.
===

• The average age and the credit score of active customers is significantly different.

• For the other variables, according to the t-test results, there is no significant difference.

3-Implementation of predictive models.

Befor deploying models, I'll create binary columns for each category and indicates the presence of the category with a 1 in the corresponding column. This is suitable when there is no ordinal relationship among the categories.

Also standardizing data is an important preprocessing step before applying machine learning algorithms for several reasons because often rely on numerical optimization techniques, such as gradient descent. These algorithms can be sensitive to the scale of input features. Standardizing the data ensures that all features have a mean of 0 and a standard deviation of 1, making the scale consistent across variables.

df= pd.get_dummies(df, columns=['geography'], prefix="is")
df["gender"] = df["gender"].replace({"Female":1, "Male":0})
exited = df.pop("exited")
df["exited"] = exited
df = df*1
df.head(10)

	credit_score	gender	age	tenure	balance	num_of_products	has_cr_card	is_active_member	estimated_salary	is_France	is_Germany	is_Spain	exited
0	619	1	42	2	0.00	1	1	1	101348.88	1	0	0	1
1	608	1	41	1	83807.86	1	0	1	112542.58	0	0	1	0
2	502	1	42	8	159660.80	3	1	0	113931.57	1	0	0	1
3	699	1	39	1	0.00	2	0	0	93826.63	1	0	0	0
4	850	1	43	2	125510.82	1	1	1	79084.10	0	0	1	0
5	645	0	44	8	113755.78	2	1	0	149756.71	0	0	1	1
6	822	0	50	7	0.00	2	1	1	10062.80	1	0	0	0
7	376	1	29	4	115046.74	4	1	0	119346.88	0	1	0	1
8	501	0	44	4	142051.07	2	0	1	74940.50	1	0	0	0
9	684	0	27	2	134603.88	1	1	1	71725.73	1	0	0	0

#--------------------------------------------
# Stadadizing the numrical columns
#--------------------------------------------

# Initialize the StandardScaler
scaler = StandardScaler()

for col in numeric_col:
    # Extract the selected column as a numpy array
    values = df[col].values.reshape(-1, 1)

    # Replace the original column with the standardized values
    df[col] = scaler.fit_transform(values)

# Display the DataFrame with the standardized column
df.head()

	credit_score	gender	age	tenure	balance	num_of_products	has_cr_card	is_active_member	estimated_salary	is_France	is_Germany	is_Spain	exited
0	-0.326221	1	0.293517	2	-1.225848	1	1	1	0.021886	1	0	0	1
1	-0.440036	1	0.198164	1	0.117350	1	0	1	0.216534	0	0	1	0
2	-1.536794	1	0.293517	8	1.333053	3	1	0	0.240687	1	0	0	1
3	0.501521	1	0.007457	1	-1.225848	2	0	0	-0.108918	1	0	0	0
4	2.063884	1	0.388871	2	0.785728	1	1	1	-0.365276	0	0	1	0